Video Transcript
Find the integral from negative
five to negative four of 𝑘 with respect to 𝑥 plus the integral from eight to
negative five of 𝑘 with respect to 𝑥, given that 𝑘 is a constant.
In this question, we are given two
definite integrals and are asked to evaluate their sum. The integrand in both integrals is
a constant denoted by 𝑘. The lower and upper limits in the
first integral are negative five and negative four, respectively. And the lower and upper limits in
the second integral are eight and negative five, respectively. Note that, in the first integral,
the lower limit is less than the upper limit. And in the second integral, the
lower limit is greater than the upper limit.
In order to answer this question,
we will make use of the following property. For any constant 𝑐, the integral
from 𝑎 to 𝑏 of 𝑐 with respect to 𝑥 is equal to 𝑐 times 𝑏 minus 𝑎. Recall that this property holds
whether 𝑎 is strictly less than 𝑏, 𝑎 is equal to 𝑏, or 𝑎 is strictly greater
than 𝑏. Letting 𝑎 equal negative five, 𝑏
equal negative four, and 𝑐 equal 𝑘 in this property, we find that the integral
from negative five to negative four of 𝑘 with respect to 𝑥 is equal to 𝑘
multiplied by negative four minus negative five, which simplifies to 𝑘 multiplied
by negative four plus five. This is equal to 𝑘.
Since the property holds for 𝑎 is
greater than 𝑏, letting 𝑎 equal eight, 𝑏 equal negative five, and 𝑐 equal 𝑘, we
find that the integral from eight to negative five of 𝑘 with respect to 𝑥 is equal
to 𝑘 multiplied by negative five minus eight, which simplifies to negative
13𝑘. Therefore, the integral from
negative five to negative four of 𝑘 with respect to 𝑥 plus the integral from eight
to negative five of 𝑘 with respect to 𝑥 is equal to 𝑘 minus 13𝑘, which
simplifies to negative 12𝑘. This is our final answer. However, note that the method we
have used to get to the answer is not the only method we could’ve used.
Let’s quickly go through another
method we could use to obtain the answer to the question. Recall that if we have two
integrals who have the same integrand, with the property that the upper limit of the
first integral is equal to the lower limit of the second integral, then their sum is
equal to the integral of the same integrand from the lower limit of the first
integral to the upper limit of the second integral. This property holds whether 𝑎 is
strictly less than 𝑏, 𝑎 is equal to 𝑏, or 𝑎 is strictly greater than 𝑏.
We can use this property to answer
the question. We can swap the order of the
integrals in the sum we are asked to evaluate and rewrite the sum to be evaluated as
the integral from eight to negative five of 𝑘 with respect to 𝑥 plus the integral
from negative five to negative four of 𝑘 with respect to 𝑥. Now, we can see that the upper
limit of the first integral is equal to the lower limit of the second integral. Therefore, using the property we
have just described, we can rewrite the sum to be evaluated as the integral from
eight to negative four of 𝑘 with respect to 𝑥.
Letting 𝑎 equal eight, 𝑏 equal
negative four, and 𝑐 equal 𝑘 in the first property that we described, we evaluate
this integral to be 𝑘 multiplied by negative four minus eight, which simplifies to
negative 12𝑘, which is the same as the answer that we found previously. So we have seen two methods that
can be used to find the sum of the definite integrals given to us in the
question. And they both give us an answer of
negative 12𝑘.
